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Question

$text In a Triangle end text A B C comma less than B less than 90 to the power of ring operator text and end text A D text is drawn perpendicular to end text B C text . Prove that end text A C squared equals A B squared plus B C squared minus 2 B C. B D$

hint Hint:

Hint :- Applying pythagoras theorem to both triangles ABD and ADC. find the AD2 from both equations substitute value of AD2 .Equate both the equations to get the required condition .

The correct answer is: Hence proved

Aim :- Prove that $A C squared equals A B squared plus B C squared minus 2 B C times B D$
Explanation(proof ) :-
As angle b is acute the perpendicular line AD lies inside the triangle
Applying pythagoras theorem in ΔACD ,We get
$A C squared equals A D squared plus C D squared$ — Eq1
Applying pythagoras theorem in ΔABD ,We get
$A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared$ —Eq2
Substitute Eq1 in Eq2 ,
$A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared$

$text By using end text a squared minus b squared equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis text and end text B C equals B D plus C D text we get, end text$

$not stretchy rightwards double arrow A C squared equals A B squared plus left parenthesis B D plus C D right parenthesis left parenthesis C D minus B D right parenthesis$
$open not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis left parenthesis C D plus B D right parenthesis minus 2 B D right parenthesis close parentheses left parenthesis AddCD text and subtract end text CD right parenthesis$
$not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis B C minus 2 B D right parenthesis$
$therefore A C squared equals A B squared plus B C squared minus 2 B C times B D$
Hence proved

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